# Deriving the maximum range and angle of a projectile

In a thread in the Green group, AnarchistAsian and I were discussing his coil guns. I posed the question of what range he could get, and he asked me to go through the physics derivation. Here it is, simplified to require just algebra and trig.

The kinetic energy of the projectile E = ½mv^{2}, gives v = sqrt(2E/m) as the speed at launch. Let θ be the angle at which you launch (θ=0° is horizontal, θ=90° is straight up). Then you can decompose the speed into two components:

v_{h} = v cos(θ) is the horizontal speed,

v_{v} = v sin(θ) is the vertical speed.

Gravity only pulls vertically, so the projectile's vertical speed will be slowed down until it reaches its maximum altitude, then it will fall back until it hits the ground. The horizontal speed will remain constant until it hits the ground and stops. To figure out the range, you need to know the time t that the projectile flies before it hits the ground; then the range is just

R = v_{h}t

Energy conservation guarantees that it's downward speed at the end is equal to its original upward speed, just with a change of sign. That also means that the total flight time of the projectile will be half going up and half going down. Once you determine how long it takes to reach the top of its flight, you're done; just double that answer :-)

In the vertical direction, the maximum height

H = v_{v}t - ½gt^{2}

(you need calculus to derive this result). From energy conservation, the initial kinetic energy in the vertical direction, E_{v} = ½mv_{v}^{2} must equal the potential energy at the top of the flight, P_{v} = mgH:

mgH = ½mv_{v}^{2}

H = ½v_{v}^{2}/g

Substitute this on the left hand side of the trajectory expression,

½v_{v}^{2}/g = v_{v}t - ½gt^{2}

v_{v}^{2} = 2gv_{v}t - g^{2}t^{2}

v_{v}^{2} - 2gv_{v}t + g^{2}t^{2} = 0

(v_{v} - gt)^{2} = 0

v_{v} = gt

So, t = v_{v}/g = v sin(θ)/g is the time to reach maximum height. Double that as discussed above, and you get the range R = 2 sin(θ) cos(θ) v/g.

Work out the angle that gets you maximum range by just plugging in different angle values and finding the one that is biggest.

Not meaning to interupt nor disrupt, but I am having a little difficulting understanding the following portion of the equation: (Y=0

Â° is horizontal, Y=90Â° is straight up).Select as Best AnswerUndo Best Answer

Ignore the italic "angstrom" symbol :-( I'bles WikiFormatting processor has some stupid problems with the extended ASCII characters (character codes 128 to 255). It sticks in extra symbols, and then if you try to edit the text to get rid of them, it actually adds even more of them!

Those two statements are supposed to be "Y = 0 degrees is horizontal, Y = 90 degrees is straight up", where I'm using "Y" as the symbol for angle instead of Greek "theta".

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becomes

C

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Θ / θ . hmmmmm works for me. ;) . #920 and #952 . . Found that by Googling "HTML +theta. Seems I have a lot more stuff to add to my formatting topic. :)

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Thanks. Those are standard Unicode values? You should be wary before going down the road of adding all of them -- there are 65,536 possible! Then there's Big 5 and the other asian fonts...

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> Those are standard Unicode values?

. I dunno. Found 'em at Entities for Symbols and Greek Letters. That page says "Glyphs of the characters are available at the Unicode Consortium," so I'm guessing they are.

.

> ... there are 65,536 possible!

. What do you think about having a spreadsheet for download? Using Fill and C&P, it shouldn't take too long to build - that's what I did for the ASCII chart screen captures.

. I'm guessing that a LOT of the characters are not needed (eg, any pictograph language chars). For use on Ibles, just the English (Roman?), Greek, and Math chars should be enough. Probably ought to include any German, French, &c chars for those times one wants to spell Gotterdammerung or tete-a-tete properly.

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German, French and Spanish are already part of the ISO Latin-1 encoding (the table you have :-). They all have proper entity names, as well as numeric values: é = é ö = ö and so on.

It's the non-Latin characters for which you have to go to Unicode (Greek, Cyrillic, the Cyrillic-derived Eastern Europeans, etc.).

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. Does that mean that all I need is a supplement to the ASCII/Latin-1 chart, with the Greek Unicode characters? Or do you think there is more that should be added? . I don't want to be Anglo-centric, but the charts may be too big if the "minor" (eg, Cyrillic) languages are included. With a SS, it wouldn't be a problem. Arrrrgggghhhh! We need tables in the markup! Then the user could copy the entity directly from the page. . heehee Yeah, Latin. Not sure where I got Roman from.

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I wouldn't do it as a supplement -- that's an open ended problem. I would make mention of the extended characters, and provide a link to a definitive site (as you provided above). The alphabet is "Roman", the character set is ISO Latin-1 :-)

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Could you add that link to your FAQ? Thanks!

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. Geez! I guess I need to stop procrastinating and make the topic into an iBle. Then I can make you a collaborator and YOU can add stuff. ;) . BTW, I added the link about half an hour ago. Right after my 10:58 PM post.

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Thanks to both GH and NM. I've tweaked the symbology in the text.

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. Considering the tools you have to work with, I think you did an exceptional job. The fractions are real crappy (probably work better at a larger text-size) and the superscripts look too big to me, but that's not your fault. . Now let's see what you can do with summation. heehee

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Thanks

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PS: you can edit any additional symbols out by moving the cursor to the right of them and backspacing over them.

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ok here is a flip I have my range and velocity of projectile. How do I get the angle that reached that distance. I also have a friction factor.

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Friction complicates matters, as you now have to solve an integral functional equation. Without the friction, just do some algebra on the expressions above; the solution is trivial, so it is left as an exercise for the reader.

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hey Kelsey! Me and my dad finished soldering parts onto a breadboard for the DC to DC boost converter needed to charge the capacitors! woo woo!!!!!!!!!!!!!

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Rock on! Congratulations on this milestone :-) You've got it troubleshot (troubleshooted? whatever...) and ready to go? I look forward to seeing your results.

If you really can get an altitude of several kilometers (I

highlyrecommend launching out to sea, if you can :-), that will be quite an achievement.Select as Best AnswerUndo Best Answer

well, the parts are on the breadboard, now we just need to make the connections...

The ible is here, it's not very detailed, but the diagram is pretty good.

altitude of several kilometers... Yeah, that makes sense i guess, since a .22 can go pretty far, even if in-accurate, but the thing is, the efficiency of the coil gun...

My dad has some 10awg wire at his lab, enough for me to have several stage's worth of wire... how many stages do you think it would take to get 75% efficiency or higher?

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Ack. Now you're getting into electrical engineering, and that's way outside my experience. It seems to me that you have the knowledge to at least search for that sort of information, either here on I'bles, or to recognize "truthy" :-) Google results. Binary guy, westfl, GH, NM, any of those folks probably know more about practical electrical stuff than I do.

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oh, ok, i see, uhh, there aren't many people on this site who know that much about coil guns specifically... There is one, but he's not very active...

Oh, right, yeah, i forgot

yourphysics andmy dad'sphysics are different...He's nano, you're high energy...

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Yes, that's true. I can do the basic calculations, and make simplifying assumptions so I can "figure out" how something works, but I don't know electrical stuff at the engineering level. My HEP work is on the detector side, and on data analysis. If I were directly involved with the accelerators, I'd probably be more familiar with projects like yours.

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Ah ha, i see, well, i guess we'll just experiment and see how many stages we can get to work...

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I highly reccomend launching it towards Mississippi. The last tornado did 25 Million dollars worth of improvement!

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xD

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Awesome! *copies for later use* On a tangent from the purpose of this thread, but how much damage would a .117" steel coated uranium BB do when it's doing about 6,000 kph?

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Well, you can do the math yourself -- you

area rocket scientist, after all ;->Uranium has a density of 19.1 g/cm

^{3}(you can probably neglect the steel coating). Your BB has a diameter of 0.3 cm. Now you have the mass and velocity, so you can calculate the kinetic energy.Use Wikipedia to look up the physical properties of the material you're shooting at (glass, wood, steel, people). That should give you enough information to convert energy to penetration depth.

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oh... i haven't learned calculus yet, let me show this to my dad...

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Please do. The only place where calculus is really needed is to

derivethe projectile height as a function of time.If you're willing to accept that equation "on faith" :-) then everything else is just algebra, plus substituting in the numerical values for your coil gun.

I would definitely encourage you to go through this with your Dad. Try it on your own first, making sure you understand each step; feel free to ask any questions you want here. If you think you've found a mistake,

definitelypost it here -- just because I wrote it down doesn't make it true :-)Select as Best AnswerUndo Best Answer

ah, yeah, he saw it, he says there are no mistakes =-), but, i still don't know trigometry, and he'll have to show me first, i'll get back to you later...

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Heh. Okay. I guess the two equations for v

_{h}and v_{v}you can "take on faith" for now. Your calculator will have buttons labelled "sin" and "cos", which you can use to plug in different angles (Y values) at the end of my writeup. You can explore the range as a function of those angles and see where it is maximized.When you take trig (or look it up on Wikipedia) you will see why those two equations work the way they do.

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calculators? no, we always do by hand, when possible...Anyways, my dad just finished showing me how to do it, (after waiting for him, and then looking all over the place for those dry-erase markers) on a dry erase board we got a long time ago...

Well, here's what i got:

^{and they are}WAY^{off, since the air resistance isn't acounted for...}Distance up and down is 50,000 meters, =-o

^{so then range is 100 km?}And... as for the angle, 45 is always the way to go, right? i thought so, and my dad said so...

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Wow, that is a

hugeamount of kinetic energy you're going to put out. With a vertical height (at 45°?) of 50 km, your half flight-time is 101 seconds, at an initial vertical speed of 990 m/s -- is that what you got? Your total time downrange is 202 s, and at 45° your horizontal and vertial velocities are equal, which means a range of nearly 200 km.With those numbers, all of my approximations are

veryapproximate :-) You not only have to account for air resistance, you have to account for thedifferentair resistance at 50 km altitude (that's halfway to low orbit!), and you probably need to take account of the Earth's curvature to compute the true downrange distance.The 45° angle is the optimal solution for maximum range. I was hoping that you'd actually go through the math -- use the range formula at the end, plug in different angles to compute sin(Y)cos(Y), maybe even make yourself a graph of range vs. angle. Then you could see the 45° solution instead of just being told :-)

This was a really good exercise; thank you! I hope that you got something out of it.

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oh, we got flight time of 100 seconds, coz we rounded.

^{less work, fairly accurate right?}990 m/s,uhh, well that's about twice what it was. going to it's highest point, it goes 25,000 M.

I calculated about 569m/s. HOPEFULLY *crosses fingers*, we can get %70 of that speed...

^{still a lot, but i have a "need for speed"}50km is way off though, the air resistance would lower that more than 10 fold... =-(

Well, we already concluded 45° was the best angle, but he still showed me how to calculate it. So it was 770m/s in that direction... But that was with 1000m/s, that i

thoughtwas the speed, but it was a mis-calculation =-(ok, so anyways, the capacitor bank has 4800uf, at 450 volts.

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Rounding off is perfectly correct, AA. I do it all the time, and it is an excellent tool for you to learn. We call it an "order of magnitude" or "back of the envelope" estimate. If I were doing this in my head, or pencil and paper, I'd use 10 m/s

^{2}for gravity instead of 9.8, and so on.About the 990 m/s vertical I asked about, maybe I got it wrong? It sounds like one of us had an extra factor of two floating around :-)

In any event, that is one heck of a gun you want to build. 972 joules of stored energy (U = ½CV

^{2}); call it 1 kJ. If you're estimating 1 km/s launch speed, I guess your projectile is only going to weigh about 2 grams? That's not very big; steel has a density of about 8 g/cm^{3}. You're going to launch something like a ball bearing?Select as Best AnswerUndo Best Answer

hmmm... 990, maybe you used the 50,000 as to the top, instead of up and back, i think that's the most likely...

close, not 2 grams, 3 grams. It's a steel rod, 3/16 inch, or 1/4 inch, i can't decide... 1/4 will have more stopping power, i'd say... both at 1inch long.

972 joules? oh crap, i got 486... *re-calculates* i still got 486! crap, somethings wrong... ok ok, so i'm going to tell you everything i did:

i squared 450 and got 20250, then multiplied by 0.002400, and got 486.

Then i divided that by 0.0015, and got 324000, then square root was 569 m/s.

hmmmm, your answer is always twice what mine is, did i make any mistakes?

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No, you are correct. 486 J.

I'mthe one off by a factor of two :-( If you do your math right, trust it. Thank you for questioning my math.As to the first point, I did use your 50 km as the top of the arc, rather than the round trip.

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ok, cool, but ironically, we're adding the voltage together, so we have 972 joules now!

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hey, uhh... for some reason i keep getting half of what you get in calculations... also, we've decided to take the 2400uf 450v caps and combine voltage instead, so now we have twice as many joules...

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For trig, you need a calculator to find the sin or cos of a number, unless you want to use a chart.

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yeah, i used a chart thingy, my dad showed me...

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huh, i couldn't find this in the group...

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I've noticed (and other people have complained :-) that new forum topics can take time to actually appear in their Group's listing. I guess you found it from my profile page, eh?

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yeah, i just thought you were still writing and waited... then i checked your page

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Hey AA! Sorry I didn't get back with you sooner. I'll get you more info that you asked about. I've been a bit busy, but now I'm back at work, so I'll have a little more time. I'll PM you with the stuff you asked about.

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Ah, cool math! We're working on stuff

sort oflike this now in math (though the calculus stuff looks a little bit advanced).I think some one said below that air resistance isn't accounted for. How would you factor that in?

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That's where you need calculus (among other things). Air resistance is a frictional force, so it acts directly opposite the direction of motion. It's roughly proportional to velocity. What you have to do is write down a function for your velocity vs. your position along the trajectory (path length), then you add in the air resistance at that velocity acting to slow you down. Then you integrate (calculus) the whole function to work out the resulting actual trajectory. It gets pretty complex, and the text interface here couldn't handle the math notation anyway.

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Good stuff, very interesting. Do you write m/s because it's easy or because you're old-school? I can't get used to writing ms

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I use the slash notation online, for clarity. If I'm writing stuff out longhand, or using LaTeX to write an actual document, then I use standard superscripts.

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